TITLE: Olympiad inequality: is this reasoning sound?
QUESTION [6 upvotes]: I am trying to show that for $a,b,c>0,\;abc=1:$
$$\underbrace{\frac{1}{b(a+b)}+\frac{1}{c(b+c)}+\frac{1}{a(c+a)}}_{X}\geq \frac{3}{2}$$
This problem is from the Zhautykov Olympiad of 2008.
Attempt:
If $X\geq \frac{3}{2}$ is satisfied for all $a,b,c$ satisfying the initial conditions then by letting $a\mapsto b$ and $b\mapsto a$, the following inequality must also be satisfied:
$$\underbrace{\frac{1}{a(a+b)}+\frac{1}{b(b+c)}+\frac{1}{c(c+a)}}_{Y}\geq\frac{3}{2}$$
The reasoning can be reversed, thus $X\geq\frac{3}{2}\iff Y\geq\frac{3}{2}$. Now, in any case:
$$X+Y=\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\geq 3\sqrt[3]{\frac{1}{(abc)^2}}=3$$
By AM-GM. Hence either $X\geq \frac{3}{2}$ or $Y\geq \frac{3}{2}$ is true, and therefore they are both true.
Is this logic valid? It seems too easy.

REPLY [4 votes]: Write $X$ as $X=f(a,b,c)$. Then $Y=f(b,a,c)$. What you are saying first is
$${\rm if}\;\; \left(\forall a,b,c\;:\; f(a,b,c)\geq\frac32\right)\;\;{\rm then}\;\; \left(\forall a,b,c\;:\; f(b,a,c)\geq \frac 32\right) $$
which is quite correct.
Then what you show is
$$\forall a,b,c\;:\; \left( f(a,b,c)\geq\frac32\;\;{\rm or}\;\; f(b,a,c)\geq\frac32\right) $$
But here you are in trouble: one cannot interchange the $\forall$ and the "or" without any further information.